Probability of distributionsHow to find the the probability using mean and standard deviationWhat is the probability that the “smartest” of them has IQ score above 120?probability and random sampleHow do you determine which probability distribution to use?Geometric distribution and lottery problemExceedance probability of normal and logarithmic distribution by matlabCalculating probability for a normal distributionUsing the normal distribution to find the probability of getting an exact valueThe number of Bernoulli trials required to produce exactly 1 success and at least 1 failure.Find the probability of a Normal Distribution random variable
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Probability of distributions
How to find the the probability using mean and standard deviationWhat is the probability that the “smartest” of them has IQ score above 120?probability and random sampleHow do you determine which probability distribution to use?Geometric distribution and lottery problemExceedance probability of normal and logarithmic distribution by matlabCalculating probability for a normal distributionUsing the normal distribution to find the probability of getting an exact valueThe number of Bernoulli trials required to produce exactly 1 success and at least 1 failure.Find the probability of a Normal Distribution random variable
$begingroup$
A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.
However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?
Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?
probability probability-distributions
asked Mar 30 at 3:20
OlivettiOlivetti
32
$endgroup$
add a comment |
$begingroup$
A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.
However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?
Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?
probability probability-distributions
asked Mar 30 at 3:20
OlivettiOlivetti
32
$endgroup$
add a comment |
$begingroup$
A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.
However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?
Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?
probability probability-distributions
asked Mar 30 at 3:20
OlivettiOlivetti
32
$endgroup$
A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.
However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?
Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?
probability probability-distributions
probability probability-distributions
asked Mar 30 at 3:20
OlivettiOlivetti
32
asked Mar 30 at 3:20
OlivettiOlivetti
32
asked Mar 30 at 3:20
OlivettiOlivetti
32
asked Mar 30 at 3:20
OlivettiOlivetti
32
asked Mar 30 at 3:20
OlivettiOlivetti
32
32
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.
Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$
answered Mar 30 at 4:46
dnqxtdnqxt
5525
$endgroup$
$begingroup$
The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
$endgroup$
– antkam
Mar 30 at 4:50
$begingroup$
Oops, I didn't notice, thanks. Will fix presently...
$endgroup$
– dnqxt
Mar 30 at 4:51
$begingroup$
I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
$endgroup$
– antkam
Mar 30 at 5:01
$begingroup$
Moms are usually right :)
$endgroup$
– dnqxt
Mar 30 at 5:08
$begingroup$
Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
$endgroup$
– Olivetti
Apr 1 at 8:55
|
show 5 more comments
$begingroup$
Hint: You need to use both the Normal and the Binomial in different parts of the computation.
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.
Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$
answered Mar 30 at 4:46
dnqxtdnqxt
5525
$endgroup$
$begingroup$
The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
$endgroup$
– antkam
Mar 30 at 4:50
$begingroup$
Oops, I didn't notice, thanks. Will fix presently...
$endgroup$
– dnqxt
Mar 30 at 4:51
$begingroup$
I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
$endgroup$
– antkam
Mar 30 at 5:01
$begingroup$
Moms are usually right :)
$endgroup$
– dnqxt
Mar 30 at 5:08
$begingroup$
Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
$endgroup$
– Olivetti
Apr 1 at 8:55
|
show 5 more comments
$begingroup$
Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.
Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$
answered Mar 30 at 4:46
dnqxtdnqxt
5525
$endgroup$
$begingroup$
The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
$endgroup$
– antkam
Mar 30 at 4:50
$begingroup$
Oops, I didn't notice, thanks. Will fix presently...
$endgroup$
– dnqxt
Mar 30 at 4:51
$begingroup$
I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
$endgroup$
– antkam
Mar 30 at 5:01
$begingroup$
Moms are usually right :)
$endgroup$
– dnqxt
Mar 30 at 5:08
$begingroup$
Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
$endgroup$
– Olivetti
Apr 1 at 8:55
|
show 5 more comments
$begingroup$
Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.
Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$
answered Mar 30 at 4:46
dnqxtdnqxt
5525
$endgroup$
Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.
Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$
answered Mar 30 at 4:46
dnqxtdnqxt
5525
edited Mar 30 at 5:05
answered Mar 30 at 4:46
dnqxtdnqxt
5525
answered Mar 30 at 4:46
dnqxtdnqxt
5525
answered Mar 30 at 4:46
dnqxtdnqxt
5525
5525
$begingroup$
The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
$endgroup$
– antkam
Mar 30 at 4:50
$begingroup$
Oops, I didn't notice, thanks. Will fix presently...
$endgroup$
– dnqxt
Mar 30 at 4:51
$begingroup$
I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
$endgroup$
– antkam
Mar 30 at 5:01
$begingroup$
Moms are usually right :)
$endgroup$
– dnqxt
Mar 30 at 5:08
$begingroup$
Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
$endgroup$
– Olivetti
Apr 1 at 8:55
|
show 5 more comments
$begingroup$
The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
$endgroup$
– antkam
Mar 30 at 4:50
$begingroup$
Oops, I didn't notice, thanks. Will fix presently...
$endgroup$
– dnqxt
Mar 30 at 4:51
$begingroup$
I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
$endgroup$
– antkam
Mar 30 at 5:01
$begingroup$
Moms are usually right :)
$endgroup$
– dnqxt
Mar 30 at 5:08
$begingroup$
Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
$endgroup$
– Olivetti
Apr 1 at 8:55
$begingroup$
The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
$endgroup$
– antkam
Mar 30 at 4:50
$begingroup$
The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
$endgroup$
– antkam
Mar 30 at 4:50
$begingroup$
Oops, I didn't notice, thanks. Will fix presently...
$endgroup$
– dnqxt
Mar 30 at 4:51
$begingroup$
Oops, I didn't notice, thanks. Will fix presently...
$endgroup$
– dnqxt
Mar 30 at 4:51
$begingroup$
I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
$endgroup$
– antkam
Mar 30 at 5:01
$begingroup$
I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
$endgroup$
– antkam
Mar 30 at 5:01
$begingroup$
Moms are usually right :)
$endgroup$
– dnqxt
Mar 30 at 5:08
$begingroup$
Moms are usually right :)
$endgroup$
– dnqxt
Mar 30 at 5:08
$begingroup$
Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
$endgroup$
– Olivetti
Apr 1 at 8:55
$begingroup$
Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
$endgroup$
– Olivetti
Apr 1 at 8:55
|
show 5 more comments
$begingroup$
Hint: You need to use both the Normal and the Binomial in different parts of the computation.
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
$endgroup$
add a comment |
$begingroup$
Hint: You need to use both the Normal and the Binomial in different parts of the computation.
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
$endgroup$
add a comment |
$begingroup$
Hint: You need to use both the Normal and the Binomial in different parts of the computation.
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
$endgroup$
Hint: You need to use both the Normal and the Binomial in different parts of the computation.
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
answered Mar 30 at 3:39
ErtxiemErtxiem
671112
671112
add a comment |
add a comment |
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